Q on associative binary operation

[Tyler Durden]

Yeah, kinda bizarre.

There's also an ambiguity that prevents one from saying Q is associative. Is 
the table defined for both directions of *? In other words, is the table 
meant to imply values for both x*y (ie, left*top) as well as y*x (top*left)? 
For most objects x*y will not equal y*x (indeed, one may be undefined as in 
the case of matrix multiplication). Anyone remember the group theoretic term 
for these kinds of groups?

-TD


>From: Tim May <timcmay at got.net>
>To: cypherpunks at lne.com
>Subject: Re: Q on associative binary operation Date: Thu, 28 Aug 2003 
>10:41:51 -0700
>
>On Thursday, August 28, 2003, at 12:14  AM, Sarad AV wrote:
>
>>hi,
>>
>>Table shown is completed to define 'associative'
>>binary operation * on S={a,b,c,d}.
>>
>>*|a|b|c|d
>>---------
>>a|a|b|c|d
>>---------
>>b|b|a|c|d
>>---------
>>c|c|d|c|d
>>---------
>>d|d|c|c|d
>>
>>
>>The operation * is associative iff (a*b)*c=a*(b*c) for
>>all a,b,c element of set S.
>>
>>So can (a*d)*d=a*(d*d)=d considered as associative
>>over * for this case as per definition?
>>
>
>This looks like a homework assignment for a class.
>
>If a, b, c, and d are all arbitrary symbols, most substitutions (such as a 
>= 2, b = 5, etc.) certainly will _not_ give (a*d)*d=a*(d*d)=d as a true 
>statement. Only special values of a and d will make that true. (For 
>example, a = 1 and d = 1 makes (1*1)*1=1*(1*1)=1. Other values may as well. 
>Finding them is up to you.
>
>Lastly, your English is again unclear. "So can (a*d)*d=a*(d*d)=d considered 
>as associative over * for this case as per definition?" isn't a proper 
>English sentence.
>
>Why do you keep posing these problems to the list? Are they homework 
>problems? Do you think the list is just too quiet and needs ill=phrased 
>puzzlers to keep it occupied?
>
>Did you ever do the coin flip experiment we suggested you do?
>
>Are you an AI which has failed the Turing Test and escaped?
>
>
>--Tim May

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